Even Number: An integer n is said to be even if it has the form n = 2k for some integer k. That is, n is even if and only if n divisible by 2.


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1 MATH 337 Proofs Dr. Neal, WKU This entire course requires you to write proper mathematical proofs. All proofs should be written elegantly in a formal mathematical style. Complete sentences of explanation are required. Do not simply write an equation; you must explain what the equation is giving and/or why it is being used. Moreover, all equations must be properly aligned with no scratch outs. Always give a conclusion. We will begin by reviewing several standard methods of proof using the following basic definitions and using the facts that the sum and product of integers are integers. Divisibility: We say that a divides b if there exists an integer k such that b = k a. For example, 6 divides 18 because 18 = 3! 6, where k = 3 is an integer. Even Number: An integer n is said to be even if it has the form n = 2k for some integer k. That is, n is even if and only if n divisible by 2. Odd Number: An integer n is called odd if it has the form n = 2 k +1 for some integer k. Prime Number: A natural number n > 1 is said to be prime if its only positive divisors are 1 and n. If n has other positive divisors, then n is called composite. Rational Number: A real number x is called rational if x can be written as a fraction m / n, where m and n are integers with n 0. Otherwise, x is called irrational. Direct Proof Consider the statements S 1! If p, then q, and S 2! " x, p( x). At times we may try to prove that these types of statements are true. To prove S 1 directly, we assume p is true and then argue that q must also be true. To prove S 2 is true, we pick an arbitrary x and argue that property p(x) must hold. Example 1. Prove the following results directly: (a) If n is an odd integer, then n 2 +1 is even. (b) If a divides b and b divides c, then a divides c. (c) For all rational numbers x and y, the product x y is also a rational number. Proof. (a) Assume n is an odd integer. Then n = 2 k +1 for some integer k. We then have n 2 +1 = (2k +1) = 4k 2 + 4k + 2 = 2(2k 2 + 2k + 1) = 2l, where l = 2k 2 + 2k + 1 is an integer because k is an integer. Thus, n 2 +1 has the form of an even number, and so n 2 +1 is even.
2 (b) Assume a divides b and b divides c. Then b = k a for some integer k, and c = j b for some other integer j. Thus, c = j b = j (ka) = ( j k )a. Because the product of integers j k is still an integer, we have that a divides c. (c) Let x and y be rational numbers. Then x = m / n and y = j / k for some integers m, n, j, and k, with n! 0 and k! 0. Then x y = m n! j k = m j n k. The products of integers m j and n k are still integers, and n k cannot be 0 because neither n nor k is 0. Thus, x y is in the form of a rational number and therefore x y is rational. Indirect Proofs Given an implication S! p " q, its contrapositive is ~ q! ~ p, which is logically equivalent to the original implication. In order to prove that S is true, it might be easier to prove that the contrapositive is true. That is, we can assume that q is not true, and then argue that p is not true. Another common method used to prove p! q is a proof by contradiction. In this case, we assume that p is true but that q is not true. We then argue that a mathematical contradiction occurs. We conclude that q in fact must be true. Proof by Contrapositive To prove p! q, assume that q is not true, then argue that p is not true. Proof by Contradiction To prove p! q, assume that p is true but q is not true. Then argue that a mathematical contradiction occurs. Notes: (i) A logical statement is often a conjunction a! b (a and b ) or a disjunction a! b (a or b ). We apply DeMorgan s Laws to obtain the negations of these statements as follows: ~ (a! b) " ~ a # ~ b ~ (a! b) " ~ a # ~ b (ii) The negation of one or the other but not both is given by both or neither. (iii) A logical statement may be in terms of a quantifier such as for every x, property p(x ) holds. The negation is there exists an x for which p(x ) does not hold. Statement:!x, p(x ) Negation:! x, ~ p(x )
3 Example 2. Prove the following results: (a) If c is a nonzero rational number, then c! is irrational. (b) If m 3 is even, then m is even. (c) For all real numbers x and y, if x + y > 0, then x > 0 or y > 0. (d) Let n be an integer. If n 2 is divisible by 3, then n is divisible by 3. (e) Let x be a real number. If x <! for all! > 0, then x = 0. (a) Proof. (By contradiction) Let c be a nonzero rational number, and suppose that c! is rational. Then c! = m / n, where m, n are integers and n 0. Moreover, c = j / k, where j, k are integers and k 0, but also j 0 because c is nonzero. We then have! = 1 c " m n = k j " m n = k m j n. The products of integers k m and j n are still integers, and j n cannot be 0 because neither n nor j is 0. Thus,! is in the form of a rational number, which is a contradiction because! is irrational. Ergo, c! must be irrational. (b) Claim: If m 3 is even, then m is even. We shall prove the contrapositive instead; we shall assume that m is not even and show that m 3 is not even. That is, we shall assume that m is odd and show that m 3 is odd. Proof. Assume that m is odd. Then m = 2 k +1 for some integer k. Then, m 3 = (2 k + 1) 3 = 8k 3 +12k 2 + 6k + 1 = 2(4k 3 + 6k 2 + 3k) + 1 = 2l + 1, where l is the integer 4k 3 + 6k 2 + 3k. Thus, m 3 is odd because it is written in the form of an odd integer. By contrapositive, if m 3 is even, then m is even. (c) Proof. Let x and y be real numbers. Assume that x! 0 and y! 0. Then by adding, we have x + y! = 0. That is, x + y! 0. Hence, x! 0 and y! 0 implies that x + y! 0. By contrapositive, if x + y > 0, then x > 0 or y > 0.
4 (d) Claim: For an integer n, if n 2 is divisible by 3, then n is divisible by 3. Proof. Assume n is not divisible by 3. We then can apply the Fundamental Theorem of Arithmetic to write n uniquely as a product of powers of primes as follows: n = p 1 r 1! p 2 r 2!...! p k r k, where r i are natural numbers and the p i are distinct primes. Because n is not divisible by 3, then p i 3 for 1! i! k. By squaring we obtain n 2 r = p 1 r 1! p 2 r ( 2!...! p k k ) 2 2r = p 1 2r 1! p 2 2r 2!...! p k k, which is a prime factorization of n 2. By uniqueness of prime factorization, the above p i for 1! i! k are the only prime factors of n 2. Because none of the p i equals 3, then n 2 is not divisible by 3. By contrapositive, if n 2 is divisible by 3, then n must also be divisible by 3. Note: The preceding result can be generalized as follows: Let p be prime and suppose k! 2. If n k is divisible by p, then n is divisible by p. (In the proof of (d), simply replace 3 with p and replace the exponent 2 with k.) (e) Claim: If x <! for all! > 0, then x = 0. Proof. Suppose x <! for all! > 0. We must show that x = 0. So suppose that x! 0. Then x < 0 or x > 0. In either case, we have x > 0. Now let! = x / 2 > 0. For this!, we then have x > x / 2 =!, which contradicts the assumption. Thus we must have that x = 0. Note: We also could have argued by contrapositive: Assume x! 0. Then we showed that there exists an! > 0 for which x <! fails. So if x! 0, then it is not true that x <! for all! > 0. By contrapositve, if it is true that x <! for all! > 0, then x = 0. Double Implication A double implication is of the form p! q, which is read p if and only if q. This statement means p! q and q! p and both implications must be proven. Example 3. Let n be an integer. Then n is even if and only if n 2 is even.
5 Proof. Suppose first that n is even. Then n = 2k, for some integer k. So then n 2 = (2k ) 2 = 4k 2 = 2(2k 2 ). Because 2k 2 is also an integer, we see that n 2 must be even. Dr. Neal, WKU Next we must prove that if n 2 is even, then n is even. But we shall prove the contrapositive instead. So assume that n is not even, i.e., that n is odd. Then n = 2k +1 for some integer k. We then have n 2 = (2k +1) 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) +1 = 2 j +1, where j = 2k 2 + 2k is an integer. Hence, n 2 is odd; that is, n 2 is not even. We have proven that if n is not even, then n 2 is not even. By contrapositive, if n 2 is even, then n is even. From both directions, we now have that n is even if and only if n 2 is even. Note: The second direction also follows from the generalization of Part (d) using the prime p = 2. In other words, if 2 divides n 2 then 2 divides n. Theorem. 2 is irrational. Proof. Assume 2 is rational. Then we can write 2 = m / n, where m and n are integers with n 0. Furthermore, we can assume that the fraction is reduced so that m and n have no common divisors. Then by squaring the fraction, we have 2 = m2 n 2 or 2n2 = m 2. Thus, m 2 is even because it is divisible by 2. It follows that the integer m must be even (by Example 3). So we may write m = 2k for some integer k. Then 2n 2 = m 2 = (2k) 2 = 4k 2 = 2(2k 2 ), which gives n 2 = 2k 2. Thus, n 2 is even because it is divisible by 2. It follows that the integer n also must be even. So both m and n are even and therefore both are divisible by 2. But this fact contradicts the assumption that we have chosen m and n to have no common divisors. This contradiction leads us to conclude that 2 cannot be written as a fraction. Thus, 2 is irrational. Corollary. For any prime p, p is irrational.
6 Disproving by Counterexample Consider the statements S = p! q and P =!x, p(x). At times we may try to prove that these types of statements are false. To do so, we can show that their negations are true, where ~ S = p! ~ q and ~ P =! x, ~ p(x ). To show that ~ S is true, we must demonstrate a case where p is true but q is false. To show that ~ P is true, we must exhibit an x for which p(x) fails. In other words, we must come up with counterexamples that disprove the original statements S and P. Example 4. Disprove the following results by finding counterexamples: (a) If f (x) is a continuous function, then f is onetoone. (b) For all natural numbers n, if n is prime then 2 n!1 is prime. (c) For all real numbers x, 3 x > x 3. (d) For all integers m and n, if m n > 0, then m > 0 or n > 0 but not both. Solutions. (a) Let f (x) = x 2. Then f is a continuous function, but f is not onetoone (because for instance f (2) = f (!2)). Thus, the statement is disproved. (b) The result holds for the prime numbers n = 2, 3, 5, and 7. But for n = 11, we have 2 11!1 = 2047 = 23 "89, which means that 2 11!1 is composite. Thus, there exists a natural number n = 11 such that 11 is prime but 2 11!1 is not prime. (c) Consider x = 3. Then 3 x = 3 3 = x 3. So there exists a real number x for which 3 x is not greater than x 3. Hence, the statement is disproven. (d) The implication is of the form S = p! (q " r). So its negation is ~ S = p! ~ (q " r) = p! ((q! r) " (~ q! ~ r) ) So to disprove the statement, we must show that there exixt integers m and n such that p is True together with both q and r, or p is True with neither q nor r. Now consider m = 2 and n = 3 which are both integers. Then m n > 0 ( p holds), and m > 0 and n > 0 (both q and r hold). Thus, we have a counterexample that disproves S. (We also could use an example such as m =!2 and n =!3.) Note: To disprove p! (q " r), we must find a case for which p! ~ (q " r) = p! (~ q! ~ r) is True. That is, p is True, but not q and not r. To disprove p! (q " r), we must find a case for which p! ~ (q! r) = p! (~ q" ~ r) is True. That is, p is True, but either not q or not r.
7 Exercises 1. Prove the following results in a formal, elegantly written, mathematical style: (a) For all integers m and n, if m is even and n is odd, then m + n is odd. (b) Let a be an integer. If a divides b, then b is an integer and a divides b 2. (c) If a divides b and a divides c, then a divides the linear combination mb + n c for all integers m and n. (d) If x and y are rational numbers, then x + y is a rational number. (e) Let m and n be integers. If m n is even, then m is even or n is even. (f) Let m and n be integers. If m + n is odd, then m is odd or n is odd but not both. (g) Let x and y be real numbers. Then x = y if and only if x! y < " for every! > 0. (h) For every prime number p, p is irrational. 2. Disprove the following claims: (a) For all real numbers x and y, if x y > 100, then x > 10 or y 11. (b) For all rational numbers x and y, if x y is an integer, then x is an integer and y is an integer. (c) If n is an odd integer, then n is even.
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